Problem 158 - A Complete Resource Book in Chemistry for JEE Main

Question

The correct sequence which shows decreasing order of the ionic radii of the elements is (a) \(\mathrm{Al}^{3+}>\mathrm{Mg}^{2+}>\mathrm{Na}^{+}>\mathrm{F}^{-}>\mathrm{O}^{2-}\) (b) \(\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}>\mathrm{O}^{2-}>\mathrm{F}^{-}\) (c) \(\mathrm{Na}^{+}>\mathrm{F}^{-}>\mathrm{Mg}^{2+}>\mathrm{O}^{2-}>\mathrm{Al}^{3+}\) (d) \(\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}\)

Step-by-Step Solution

Answer

Option (d): \( \mathrm{O}^{2-} > \mathrm{F}^{-} > \mathrm{Na}^{+} > \mathrm{Mg}^{2+} > \mathrm{Al}^{3+} \).

Step 1: Understand Ionic Radii

Ionic radius refers to the size of an ion. Cations (positively charged ions) tend to be smaller than their parent atoms because they lose electrons, resulting in less electron-electron repulsion. Anions (negatively charged ions) tend to be larger as they gain electrons, increasing repulsion.

Step 2: Recognize Trends in Ionic Radii

In general, for isoelectronic species (ions having the same number of electrons), the ionic radius decreases with an increase in nuclear charge (more protons). The order of decreasing ionic radii for common ions (based on increasing nuclear charge) is typically: \[ \mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Na}^{+}, \mathrm{Mg}^{2+}, \mathrm{Al}^{3+} \] This sequence is due to increasing nuclear charge attracting the electrons more strongly and reducing the size of the ion.

Step 3: Analyze the Given Options

Review each option to see which matches the trend: - (a) \( \mathrm{Al}^{3+} > \mathrm{Mg}^{2+} > \mathrm{Na}^{+} > \mathrm{F}^{-} > \mathrm{O}^{2-} \) - (b) \( \mathrm{Na}^{+} > \mathrm{Mg}^{2+} > \mathrm{Al}^{3+} > \mathrm{O}^{2-} > \mathrm{F}^{-} \) - (c) \( \mathrm{Na}^{+} > \mathrm{F}^{-} > \mathrm{Mg}^{2+} > \mathrm{O}^{2-} > \mathrm{Al}^{3+} \) - (d) \( \mathrm{O}^{2-} > \mathrm{F}^{-} > \mathrm{Na}^{+} > \mathrm{Mg}^{2+} > \mathrm{Al}^{3+} \) The correct order according to known trends is (d), which matches \( \mathrm{O}^{2-} > \mathrm{F}^{-} > \mathrm{Na}^{+} > \mathrm{Mg}^{2+} > \mathrm{Al}^{3+} \).

Step 4: Conclude the Correct Sequence

Based on the analysis, the option that reflects the decreasing order of the ionic radii considering isoelectronic species is option (d) \( \mathrm{O}^{2-} > \mathrm{F}^{-} > \mathrm{Na}^{+} > \mathrm{Mg}^{2+} > \mathrm{Al}^{3+} \).

Key Concepts

Isoelectronic Species

Isoelectronic species refer to atoms or ions that have the same number of electrons. This means that although these species might be different elements or ions, their electron configurations are identical. A great example is when you consider species like - \( \mathrm{O}^{2-} \)- \( \mathrm{F}^{-} \)- \( \mathrm{Na}^{+} \)- \( \mathrm{Mg}^{2+} \)- \( \mathrm{Al}^{3+} \).
All these ions have the same number of electrons as the noble gas Neon, which is ten.
Adding or removing electrons doesn't change the total electron count in these cases, but it does affect ionic radii as nuclear charge varies. Isoelectronic species demonstrate that it's not just about the total number of electrons, but also about how those electrons are arranged around increasingly charged nuclei. Understanding this concept is crucial because even with the same number of electrons, the characteristics of these ions such as their size (ionic radii) can vary significantly depending on their nuclear charge.

Cations and Anions

Ions are charged particles formed when atoms gain or lose electrons. Cations are positively charged ions formed when an atom loses electrons, whereas anions are negatively charged ions created when an atom gains electrons. In the context of ionic radii:

• Cations: Because cations lose electrons, there are fewer electron-electron repulsions, leading to a smaller size compared to the original atom. An example is \( \mathrm{Na}^{+} \), which is smaller than a neutral sodium atom.


• Anions: Anions gain electrons, which results in increased electron-electron repulsions. This usually makes anions larger than their parent atoms. For instance, \( \mathrm{F}^{-} \) is larger than a neutral fluorine atom.

These changes in size are because the distribution of electrons alters as electrons are added or removed. It's important to note that despite having more electrons, the size change in anions also results in a thicker electron cloud, influencing their interactions within compounds.

Trends in Ionic Sizes

When analyzing ionic radii, especially among isoelectronic species, it's vital to consider the number of protons in the nucleus, otherwise known as the nuclear charge. Generally:

• For isoelectronic species, as the nuclear charge increases, the ionic radius decreases. This is due to the increased positive charge which pulls the electrons closer to the nucleus, thus reducing the size of the ion.


• For the series \( \mathrm{O}^{2-} \), \( \mathrm{F}^{-} \), \( \mathrm{Na}^{+} \), \( \mathrm{Mg}^{2+} \), and \( \mathrm{Al}^{3+} \):
• \( \mathrm{O}^{2-} \) has the largest radius because it has the fewest protons pulling on the electron cloud.
• \( \mathrm{Al}^{3+} \) has the smallest radius, due to possessing the most protons.

Recognizing these trends is essential for predicting and understanding how ions will behave in various chemical contexts, particularly when interpreting ionic interactions and bonding.